Challenging problem given to students

Challenging problem given to students - square in a quadrant

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2 жыл бұрын

Thanks to John H. for the suggestion! A square is inscribed in a quarter circle with two of its vertices on the arc. If the radius equals 1, what is the area of the square? Special thanks this month to: Robert Zarnke, Kyle, Mike Robertson, Michael Anvari, Daniel Lewis. Thanks to all supporters on Patreon! / mindyourdecisions
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Пікірлер: 789

@EugeneKhutoryansky 2 жыл бұрын

That is thinking outside the box, but inside the circle.

@akhandanand_tripathi 2 жыл бұрын

That pun was out of the mind, but around the circle

@Zzzz-lg3iw 2 жыл бұрын

@@akhandanand_tripathi Damnnnn nice one 👍🏻

@zakthayer9315 2 жыл бұрын

inside the quarter circle*

@IS-py3dk 2 жыл бұрын

@@akhandanand_tripathi LOL

@IS-py3dk 2 жыл бұрын

This joke was 😂

@pierrecurie 2 жыл бұрын

There's an even easier way to do this. Make 3 additional copies and complete the circle. There will be 5 squares in the shape of a +. One of the diameters of the circle will intersect the outer corners of 2 outer squares. Let s = side of square. That diameter will be the hypotenuse of a right triangle (2). The legs are s & 3s. So, 10s^2 = 4 -> s^2 = 2/5.

@paulstillman4949 2 жыл бұрын

I think that's how a 12 y.o. would approach this problem

@prashantgupta263 2 жыл бұрын

i also did this with a similar approach ...but i proved the pizza shaped parts as 1/8 th of a circle with a radius = (length of square=x) ...and after that with the help of central square's diagonal i got a equation ... x+x/square root(2) = 1 after that area of square with side x came to be 0.34 but thats not the correct answer.. why??

@approximatelybalut3653 2 жыл бұрын

I did the same thing

@justpaulo 2 жыл бұрын

Same approach. You still have to argue that by symmetry the points where the square touches the "x axis" and the "y axis" are at the same distance from the circle's center.

@williamezradahab4493 2 жыл бұрын

@@justpaulo For formality, sure. In any kind of mathlete competition, though, you'd just see it's obviously symmetric and skip that step.

@no_Ray_bang 2 жыл бұрын

this was a really satisfying solution

@kyokajiro1808 2 жыл бұрын

i can't believe i managed to solve this myself before he revealed the answer

@noman3710 2 жыл бұрын

@@kyokajiro1808 nice! Congratulation! Brother

@hasindudilshan6090 2 жыл бұрын

@@kyokajiro1808 nice

@thebosstaco6870 2 жыл бұрын

haha, and wasted 30 minutes of my life to get it wrong :D

@kyokajiro1808 2 жыл бұрын

@@thebosstaco6870 bruh 30?? it's not even that hard cuz of pythagorean theorem and stuff

@vinayakraj7065 2 жыл бұрын

I went for a very simpler approach. Hear me out. The length of square’s diagonal is side*sqrt(2) and length of line between center of circle and closer square vertex used in diagonal is side/sqrt(2). These two line segments form a right angle triangle with radius as the hypotenuse. Using Pythagoras theorem, Sq(sqrt(2)*side) + Sq(side/(sqrt(2)) = Sq(r) Since r is 1 and area equals Sq(side), on simplifying equation, we get Sq( side ) * 5/2 = Sq(1) and hence Sq(side) = 2/5 Area = 2/5

@alishia7168 2 жыл бұрын

How did you find the length between the center and the closer vertex is side/sqrt(2)

@Zollaho 2 жыл бұрын

Yes I did the same. Best method. Took 1 minute to spot the method and another minute to solve the question mentally. You reach the solution directly. Area of square = 2/5. Without having to deal with half sides.

@Zollaho 2 жыл бұрын

@@alishia7168 in a 45° right triangle the side length is equal to the diagonal/sqrt(2).

@kojak8403 2 жыл бұрын

Yeah, same here

@armwrestling_nerd 2 жыл бұрын

I did the same by "cutting up the square by its diagonals". But then putting the top 2 below the bottom two to form a rectangle with sides X and 2X with diagonal 1 this ends up in 2X*X = 2/5 since knowing that X²+(2X)²=1² ==> X = 1/√5....

@danmerget 2 жыл бұрын

I used the intersecting chords theorem. My reasoning was the same until 2:40, but then I looked at the chord of length 2x, and the diameter-length chord that bisects it. The tiny line segment between the 2x chord and the circle has length 1-3x, so the rest of the diameter-length chord has length 1+3x, so the intersecting-chords theorem says x² = (1-3x)(1+3x) => x² = 1-9x² => x² = 1/10 => area = 4/10.

@dhpbear2 2 жыл бұрын

I think this solution would be above a 12-year-old's 'pay-grade' :)

@cau1834 2 жыл бұрын

Cool. I liked that

@FarihaBinteAli 2 жыл бұрын

Nice thinking, really :)

@reubenbenjamin8128 2 жыл бұрын

It's very interesting that almost all questions that arise with circles, is solved using similar triangles, even the solution you said using intersecting chords theorem is again derived from similar triangles which is the part(using similar triangles) from the question you deviated to get the same solution

@josemath6828 2 жыл бұрын

Interesante forma de resolverlo.

@markgoretsky766 2 жыл бұрын

Projections of the square vertexes form 2 equal segments on each vertical and horizontal axis. Distance from point zero to the right vertex is R=1. Applying Pythagoras gives the lengths of the segments.

@satyapalsingh4429 2 жыл бұрын

My heart is filled with joy by watching your video .Method of solving is praiseworthy .Keep it up , dear professor ! My salutations to you !

@lphenry1 2 жыл бұрын

I Feel like the most direct way is apply Pythagoras theoRem twice : - once on the bottom left triangle with sides a, a, x -> 2a^2=x^2 (1) - once on the right triangle formed by the center of the circle and the top and bottom corners of the square, of sides a, sqrt(2)x, 1 -> a^2+2x^2=1 (2) 2x(2)-(1) -> 4x^2=2-x^2 -> x^2=2/5

@studyholicandgodsservant753 2 жыл бұрын

Yes dude I was also thinking the same this is an alternative and easy method I apply for most of the geometry questions.

@cindycster1 2 жыл бұрын

Yup! That’s what I did too

@duartesilva7907 2 жыл бұрын

Agree.

@HBSuccess 2 жыл бұрын

Yup. MeeToo.

@LeftGuard 2 жыл бұрын

Others have mentioned it, but the easiest way is to complete the circle which produces a cross made of five identical squares, each with side length x. Drawing a diagonal across a three square rectangle gives you a diameter, and the equation x² + (3x)² = 2² drops straight out of it, meaning 10x² = 4, so x² (the area we are looking for) = ⅖.

@aarondriscoll287 2 жыл бұрын

This is the method I used, as well. Very quick to do in your head

@pierrecurie 2 жыл бұрын

I figured I wasn't the first lol

@dianeweiss4562 2 жыл бұрын

I really prefer the simplicity of this solution.

@martinpaddle 2 жыл бұрын

Even simpler is to notice that the area of the square is the same as the area of the rectangle with diagonal going from the center of the circle to the right-most vertex of the square (for example, by dissecting the square into four equal triangles and rearranging these into said rectangle). The diagonal of this rectangle has length 1, and the base is twice the height, which gives all the information needed.

@armwrestling_nerd 2 жыл бұрын

Not sure that's the easiest one that you suggested... Cutting up the square in 4 triangles from its diagonals putting the top 2 below the other ones creates a rectangle with diagonal 1 and sides x and 2x. Which more or less by sheer ocular inspection reveals its solution 0.4 ...at least for me

@bificommander7472 2 жыл бұрын

I did it slightly different. I looked at the diagonal d of the square. The top point on the circle is then 0.5d to the right and d to the top of the origin of the quarter circle. This gives a right triangle whose diagonal side is the radius of the circle so 1^2=(0.5d)^2+(d)^2 = 1.25 d^2 The area of the square is 0.5d^2, which is 2/5 times the value of the radius of the circle we found before, so the area is 2/5*1=0.4

@bobtivnan 2 жыл бұрын

That's how I did it. Seems a bit less complicated.

@derrickthewhite1 2 жыл бұрын

That was my method. Throw it on a grid!

@MrPLC999 2 жыл бұрын

The average 12 year old will not be able to even understand the problem, much less solve it.

@kanishkmukherjee08 2 жыл бұрын

This one would've been a nightmare for me and my homies had it been given to use before you uploaded

@hedgehogclaws8877 2 жыл бұрын

Before watching or reading comments, I have a few things to say. First, I constructed the whole circle by adding 3 more quarter circles and realizing the square in the center is the same as the square we’re trying to find the are of. Since the square on the opposite side of the square shown must be the same as the square shown, I can connect these 3 squares into a rectangle with 2 as the length of the diagonal because the diagonal of this rectangle is the diameter of the circle. Next I can use the pythagorean theorem to construct this equation 2²=x²+(3x)² 4=10x² ²/₅=x² I don’t have to go further since x² will represent the area of the square, so the answer is 0.4. The thing that interests me however is that this problem didn’t feel as difficult as problems with similar titles. Like I perhaps could’ve solved this when I was 12, so I’m wondering if anyone else thinks similarly. Edit: just watched the video, no WAY I would’ve come up with that. Elegant solution, and indeed difficult! (At least for me that is)

@hedgehogclaws8877 2 жыл бұрын

@UCIyRFlOxrWvPQES6NbjHWCQ I didn’t explain very well in my comment, but I can prove what I mean. It’s difficult for me to type it out right now, but other commenters had the same solution so if you can find their comments they can offer more clarity

@Tim3.14 2 жыл бұрын

I like your method a lot. If I may summarize: Duplicate the figure to complete the circle. Add a diameter going between opposite corners of 2 squares. This is a right triangle with s^2 + (3s)^2 = 2^2. Solve for s^2 and you're done! 😊

@Tim3.14 2 жыл бұрын

@Visi0ons That leg of the triangle is from the sides of 3 congruent squares (including the square formed at the center of the circle).

@cau1834 2 жыл бұрын

I connected the top vertice of the first square to the bottom vertice of the of square under the fist, linking two diagonals. Then I connect the top vertice of the first square to the top vertice of the square on the left, creating a segment equal to two half diagonals. That formed a right triangle with the hypotenuse measuring 2 (diameter). Then I used Pythagoras.

@monicaecheverria1026 2 жыл бұрын

Yo también construí el círculo entero, y obtuve 7 cuadrados iniciales, todos dentro de un octógono. Con senos y cosenos obtuve las medidas del triángulo del octógono. Calculé su área y dividí para 7. Esto a los 12, no hubiese sido posible.

@loganv0410 2 жыл бұрын

Coordinate system approach: Beginning with the 45-degree angles at 2:00 and the square's side length s, set the origin at the center of the circle. The right-hand point of the square is at coordinates X equals the square's diagonal and Y equals half the diagonal. By Pythagoras r^2 = 1 = (5/2) s^2 or s^2 = 2/5.

@luke9947 2 жыл бұрын

Take either one of the two points of the square that are on the circumference and project it on both x and y axis. You can easily find the length of those projections (call them A and B) as functions of the side of the square. Then you do the pythagorean theorem with the radius of the circle as the Hypotenuse and A and B as legs.

@dermitdog1699 2 жыл бұрын

I used two quarter-ellipses of dimensions 1,1/2 and 1/2,1 to solve this. Their intersection point is the centre of the square, and then it's a matter of using Pythagoras to find the side lengths and squaring it.

@wortexTM 2 жыл бұрын

I found a completely different (slightly more 12yo) approach to this. Draw a full circle with all 4 quarters like this one. This creates an additional square in the middle that's identical to the 4 others. Take 3 of those squares in a line now, assuming the original's square side was x, three squares will create a 3x by x rectangle with 2*radius as its diagonal. We know the radius is 1 so now just apply Pythagorean theorem et voilà! You get the x^2 equal to 2/5

@hedgehogclaws8877 2 жыл бұрын

This was my solution as well

@shaswatadutta4451 2 жыл бұрын

Your solution is very creative.. really appreciate it!!

@No_king1143 2 жыл бұрын

What does Pythagorean theorem etc voilá means?

@jacoboribilik3253 2 жыл бұрын

how do you know the diagonal passes through the center?

@astrid4815 2 жыл бұрын

Nice! I love your math videos! Keep it up!

@leejamison8436 2 жыл бұрын

I worked it out similarly, but extending the problem out to the full circle, realizing there would also be a square matching this one on the opposite quadrant, which would necessitate a square centered at the axis of the circle as well. I then drew a diagonal of length 2r across the three squares creating a hypotenuse. Applying the Pythagorean Theorem it was clear the hypotenuse, squared to four, would have been the product of adding a square of three units to a square of one unit to produce ten units with a total area of four. So the tenth part of that would be an area of .4. Thus I worked out the area of the square without ever actually pinning down the lengths of any of the sides other than the radius and the circumference.

@koenth2359 2 жыл бұрын

Starting with an upscaled version: Centre at origin and two of the square's vertices are at coordinates (1,0) and (2,1). The last vertex is on the circle, which by Pthagoras has radius sqrt(5). At this scale the square has area 2. Now to scale down to the original size, we divide all distances by sqrt(5), areas will scale down by the square of that, so 5. Answer: square area = 2/5.

@MrSaemichlaus 2 жыл бұрын

I drew a line from the center of the circle to one corner of the square that touches the circle. I calculated the length L of that line parametrically from the side length S of the square, as the pythagorean sum of the cartesian components of the line. Set the length L equal to 1 which is the radius of the circle, and you can solve for S and then S^2, the area of the square.

@amuhnra3391 2 жыл бұрын

MUCH easier method: Call the vertical diagonal of the square 2a, draw a triangle from the ends of that line to the circle's center. This is a right angled triangle with side lengths a, 2a, and 1, where 1 is the hypotenuse. a²+(2a)²=1 so (2a)²=0.8 This is the area of the square of the diagonal, so we need to halve it to get the area of the original square. 0.4 (using the fact that for any square of area N, if you take the square of the diagonal, you get a square with area 2N)

@Tiqerboy 2 жыл бұрын

Yeah, I more or less solved it that way. I called the length of the side of the square a, so its area is a². In my case, the lengths of that triangle you describe are a√2 (diagonal of the square), a/√2 and 1. By Pythagorean theorem you have 2a² + a²/2 = 1, solving for a² gives you 2/5, and that's the answer. What surprised me is that Presh needed three twitter followers to help solve that problem. This should be one he'd easily get on his own.

@arkamitra5354 2 жыл бұрын

That's a great alternative approach. Correct me if I am wrong, but your method ASSUMES that the diagonals of the square will be parallel to the two straight sides of the quarter circle (and hence the right angled triangle that you made) Sure, the assumption ends up being true because intuition alone will tell you that the symmetricity of a square will make it impossible to have the configuration of the square diagonals in any other way (than the one assumed), but still a proper solution cannot be based of any assumptions/intuitions. And if we do work off this assumption, we get the 45° angles pretty easily, and then Presh's method is very very similar to yours, can't call one easier than the other.

@patrickng8974 2 жыл бұрын

same approach as I did

@arkamitra5354 2 жыл бұрын

@@Tiqerboy I think proving the symmetry (parallelism of the diagonals in your case) is the more difficult part of the question, which is why Presh may have needed help. Picking a triangle to apply Pythagoras is the easier part :)

@Tiqerboy 2 жыл бұрын

@@arkamitra5354 yes, I proved it in my head that the bottom left triangle is a right angle 45° one in the diagram, the intersections of two vertices of the square and quarter-circle, i.e. with the horizontal and vertical axes, have to be equal in length by symmetry, so I went on to solve it from there. Yes, I skipped some formal steps, which I suppose is why i lost interest in formal mathematics years ago.

@EllipticGeometry 2 жыл бұрын

The topmost vertex of the square is at a multiple of Cartesian coordinates (1, 2). Divide that by its length of sqrt(5) to match the radius of 1. The diagonal of the square is therefore 2/sqrt(5). Divide by sqrt(2) to obtain side length sqrt(2/5), which when squared gives area 2/5.

@mike1024. 2 жыл бұрын

Clever! I thought about that bisector and the smaller triangles and even the segment that ended up being the hypotenuse of your large triangle. I didn't quite put it all together without writing it down. I kind of expected pi to be in the answer.

@mathisnotforthefaintofheart 2 жыл бұрын

it's fascinating that there are so many different ways of solving this problem

@trnfncb11 2 жыл бұрын

Once you've established the symmetry of the problem (i.e. basically that the two diagonals of the square are parallel to the two radii delimiting the quarter circle), it is somewhat simpler. Denote by O the center of the circle and by A, B, C, D the corners of the square (clockwise starting from the leftmost). Draw the vertical from C to intersect the lower radius at point F and set OD=CF=DF=a. Then (i) the side of the square is a*sqrt(2) and (ii) OCF is a right triangle with legs equal to a and 2a and hypotenuse equal to the radius r (=1). So r^2=5a^2 and the area of the square is 2a^2=2r^2/5.

@knexfan0011 Жыл бұрын

You can also use trig: the point at which the circle touches the top corner of the square is twice as far up (1 diagonal of square) as it is off the side (1/2 diagonal of square). Draw a line from that point to the origin. This line forms an angle a with the vertical line. That angle combined with the relationship established in the beginning can be used to write sin(a) = 2cos(a). Solve for angle a = atan(0.5). The diagonal of the square is cos(a). Based on that, you can calculate the area as cos(a)²/2.

@patsk8872 2 жыл бұрын

I got 2/5 quickly using this method: the right triangle must be symmetrical and thus a 45-45-90 right triangle. Complete the circle along with an identical square in the other 3 quadrants. If the square has side length "s" you will be able to mark out all segments in terms of s. You can find an intersection of chords at the bottom of the square in the video. Applying the formula a*d = b*c for intersecting chords you'll be able to get the equation 2s^2 = (1-s/root2)(1+s/root2), leading to s^2 = 2/5 = area of the square.

@ps.2 2 жыл бұрын

Props for solving a problem using the Gougu theorem while managing to avoid mentioning it by name. I saw what you did there.

@davidseed2939 Жыл бұрын

Note by symmetry, the rt. Δ is isosceles ie 45°, let x be the projection of the side the square onto r. the hypotenuse is the side of the square length sqrt(2)x and so area of square is 2x² consider the lower point of square touching circle. horizontal distance 2x, vertical distance x, diagonal distance sqrt(x² + (2x)²) = sqrt(5)x r=¥5.x=1 , 5x²=1, x²=1/5, Area=2x²=0.4

@bpark10001 9 ай бұрын

You can also use triangle formed by vertical diagonal of square, bottom left part of radius & radius of circle to the uppermost vertex of square. Equation is R^2 = [sqrt(2)*X]^2 + [X/sqrt(2)}^2. 1 = X^2(2 + 1/2) area = X^2 = 2/5

@Fisherman-pe1ew 2 жыл бұрын

I solved this by using coordinate points and Pythagorean. I said the x intercept was (x,0) and the y intercept (0,y) which means the two points where the circle intersects the square are (x,2y) and (2x,y). Since the radius is always 1, I set up a system of equations using Pythagorean to get x^2 + (2y)^2 = 1 and (2x)^2 + y^2 = 1. By using substitution, you can find X (or y) to equal (rt5)/5. You can use Pythagorean again using the small triangle in the bottom left since X^2 + Y^2 = side^2, and solve that for a side length of (rt10)/5. Then you just square that to get 10/25 or 2/5 for the area.

@davidp4427 2 жыл бұрын

I just drew a radius to the top vertex of the square and dropped a line down through the diagonal of the square. It was then apparent that the resulting right triangle had sides equal to the diagonal of tbe square, one half the diagonal of the square, and one. Apply the Pythagorean theorem and solve for the length of the diagonal. From that you can get the square's sides and the area.

@williamharrison5387 2 жыл бұрын

There's an easier way to do this. Construct a line from the center of the circle to the top of the square. Draw a line down from where this line intersects the circle until it intersects the x-axis at a 90° angle. The base of this newly-constructed right triangle is half the diagonal, and the height is the diagonal. The hypotenuse of the triangle has length 1 (it is a radius of the circle) so the sum of the squares of the legs must equal one. Let the base of the triangle be x. x^2 + (2x)^2 = 5x^2 = 1, thus x^2 = 1/5, x = 1/sqrt(5), and the base and height of this triangle are 1/sqrt(5) and 2/sqrt(5). The area of the square is just the product of these (diagonal * 1/2(diagonal) = area (for a square)) so it is 2/5 = 0.5

@wwoods66 2 жыл бұрын

"2/5 = 0.5"; typo: 2/5 = 4/10 = 0.4

@vaibhavdalmia3623 2 жыл бұрын

I did this in the same way you did. Yeah this is more easier.

@Akronox 2 жыл бұрын

Yes this is much faster, it is even faster if you don't solve for x. I just solved the diagonal squared (D² + (D/2)² = 1² => D² = 0.8) and then we know that since a diagonal = sqrt(2) * the side of the square, the area of the square is equal to D²/2 = 0.4.

@gm_matthew 2 жыл бұрын

Yes, this is how I did it too.

@SenselessUsername 2 жыл бұрын

Alternatively: From symmetry the four corners have coordinates (x,0), (0,x), (x,2x) and (2x,x) for some x, so the surface of the square is 2xˆ2. HOwever the last two coordinates are on the unit circle (equation: 1 = X*X + Y*Y), so 1= 4xˆ2 + xˆ2 = 5xˆ2. This means xˆ2= 1/5, and the surface is 2/5.

@antonyqueen6512 2 жыл бұрын

You have an error. The surface is 4x^2, not 2x^2. I did a similar approach using the symmetry of the construction and calculated the side L of the square directly by substituting x=L.sin(45)=L.sqrt(2)/2 ==> (using point (2x,x) on circle) 2L^2 + 1/2.L^2=1 ==> Area A A= L^2 = 4/5

@michaelkouzmin281 2 жыл бұрын

I used analitical geometry. Let d = diagonal of the square. Then coordinates of the uppermost angle of the square will be x=d/2; y=d; coordinates of the rightmost angle are x=d; y=d/2. Both points belongs to unit cicumference, so x^2+y^2=1 => d^2+ (d/2)^2=1=> d^2=4/5. A square is a rhomb , so it area = d^2/2=2/5.

@jeper3460 2 жыл бұрын

the radius of the circle sector is 1. draw a line from the corner to where the square meets the perimeter of the circle sector. the short leg has a length equal to half the diagonal of the square, call that length x. the longer leg has a length equal to the diagonal of the square 2x. sqrt(x^2+(2x)^2) = 1 => sqrt(5x^2) = 1 => 5x^2 = 1 => x = 1/sqrt(5) the diagonal of the square is 2x = 2/sqrt(5) the diagonal is the hypotenuse of a 45/45/90 triangle with the legs being two of the sides of the square. divide 2x by sqrt(2) to get the length of the square which is equal to sqrt(2/5) the area of the square is one of the sides squared which is sqrt(2/5)^2 = 2/5. The area is 2/5

@ArchitectMouaed 2 жыл бұрын

My heart jumps with joy when I see the video

@adamyaser7136 Жыл бұрын

Analysis way: the function of the half circle with radius 1 would be : y=/(1-x^2) (y=/r^2-x^2 in the general). You put 2x for y, because you want the the value of x, that would give you 2x and would be on the circle. You solve the equation. And then you calculate: x^2 + x^2 to get the hypotenuse^2, which is the side of the square ^2, which is the area of the square

@MorswinCzcigodny 2 жыл бұрын

You can simply draw a line from the center of the circle to the uppermost corner of the rectangle and then to the lowermost corner. This gives you the right riangle which sides are (d/2 ; d ; 1) where d is the diagonal. Solve for d^2 which is d^2 = 4/5. Area of a rectangle is a half of that: A=d^2/2 = 0.4.

@mahinnazu5455 2 жыл бұрын

U r my best teacher over the world.. So I pray for u always...

@nathangallagher5112 2 жыл бұрын

My method was to just use the equation for a circle. I drew a line through the horizontal diagonal of the square, and focused on this lower part of the quarter circle. After some basic geometry relations, you can figure that you need the point where x=2y for the equation of a circle. Just plug in (2y)^2 +y^2 = 1^2, and you will get y=1/sqrt (5)=sqrt(2)/2*s. And then solve for s, the side length of the square, which you can square for the area.

@siddharthsabharishankar7714 2 жыл бұрын

I connected the centre of the quarter circle and one of the points where the square touches the circle. Taking the side of the square to be x, I can form a right angled triangle with base as x/root2, xroot2 and hypotenuse as 1. On solving, x^2= 0.4 which si the area of the square. Beautiful!

@ps.2 2 жыл бұрын

Yep, that's how I did it. Though you do have to first satisfy yourself that the problem is symmetric so that the angles are 45°. I did that by inspection/intuition, but it would take a bit of effort to actually prove it.

@mmattoso1 2 жыл бұрын

I did as following: by simmetry, the side of the square is placed with 45° inclination in relation to the horizontal and to the vertical lines. By using congruence of triangles, it is easy to prove that the diagonal of the square is perpendicular to the "floor" (horizontal line). Then we call x the square side and just use Pitagoras on the triangle made by the center of the circle, the bottom vertex and the top vertex of the square: the hypotenuse will be the radius 1 and the cathetus will be x.cos(45) = (1/2).x.sqrt(2), and the square diagonal x.sqrt(2) => 1 = x^2/2 + 2.x^2 = 5.x^2/2 => x^2 = square area = 2/5

@OrenLikes 2 жыл бұрын

I went with height (=base), h, of the triangle T at the center of the circle. T is a quarter of the square. The diagonals of the square are 2h. We get a right triangle of h, 2h, and 1 at either contact of the square with the circle. h²+(2h)²=1 -> 5h²=1 -> h²=1/5 = 2T = 2/4 of the square's area. Square's area = 4T = 2/5 (=0.4).

@jpolanco2004 2 жыл бұрын

Law of cosines: there is a triangle with sides x, x/sqrt(2), and 1. The first 2 form an angle of 135 degrees, and x is the length of the desired square. From there, we get 1 = x^2[1/2 + 1 + 1], so x^2 = 2/5.

@AvoidsPikes- 2 жыл бұрын

I was thinking about that too, but I don't think that we covered the Law of Cosines in Geometry at this point.

@johnspathonis1078 2 жыл бұрын

Yep!! That was the way I did it. Problem is I don't think 12 years old children learn the cosine rule.

@jpolanco2004 2 жыл бұрын

Law of Cosines comes from Pythagorean Theorem, so we just add those extra steps.

@mathisnotforthefaintofheart 2 жыл бұрын

@@johnspathonis1078 In China I am sure they do. These kids don't have time to play Fortnight or waste their time listening to Taylor Swift.

@jackburns6403 2 жыл бұрын

@@mathisnotforthefaintofheart I like the implication that Taylor swift has single handedly impacted a generations intelligence

@erikcolban8268 2 жыл бұрын

Let O be the center of the circle, A the bottom vertex of the square, B the top vertex of the square, and x half the diagonal of the square. The triangle OAB is a right triangle with sides x, 2x and 1. Therefore, x^2 + 4x^2 = 1, thus x^2 = 1/5. The square of the side of the square is x^2 + x^2 = 2/5.

@rinkuanupambhadra6064 2 жыл бұрын

Another problem that is solvable, to boost our confidence

@AvoidsPikes- 2 жыл бұрын

Or make me feel like a big dummy 😂

@martinpaddle 2 жыл бұрын

The area of the square is the same as the area of the rectangle with diagonal going from the center of the circle to the right-most vertex of the square. This can be seen, for example, by dissecting the square into equal triangles and rearranging these. The next thing to notice is that the diagonal of this rectangle has length 1, and the base is twice the height. Using Pythagoras you get the base and height (h=1/sqrt(5)), and from this that the area is 2/5 (haven't watched the video)

@martinpaddle 2 жыл бұрын

edit: just watched the video, they make it more complicated than necessary

@Maxime7101 2 жыл бұрын

Per symmetry, the angles are 45deg. If we take the intersection point between the square and the circle, its coordinates have to be x=2*sq(2)/2 * c and y=sq(2)/2 * c. But because the point is on the circle, we have x^2+y^2=1 and at the end, we find c^2=2/5

@jeffmartin5419 Жыл бұрын

I did a variant: once I figured out the inner triangle must be 45-45-90 (roughly the way you did), I labeled the sides (center of circle to corner o the square) n. (x is a side of the square.) Then I drew a line strait down from the rightmost corner of the square tangent to the circle - which has length n. The distance from the center to where that line hits the x-axis is the same as the diagonal of the square, which is also 2n. So I have a right triangle with a hypotenuse 1 and side lengths 2n and n. From there, I also note that 2n = x(sqrt 2), and solve for x ^2.

@richardgurney1844 2 жыл бұрын

That solution is so much simpler than what I ended up doing! I labeled the side of the square as X Since X is the hypotenuse of the 90,45,45 triangle by the circle's centre, by the Pythagorean Theorem the side of this triangle is Xsqrt(2)/2 - in other words the square's lower vertex touching the arc is at a height of Xsqrt(2)/2 above the radius, so by definition Sin(theta) = Xsqrt(2)/2 The diagonal the square, by Pythagorean Theorem, is Xsqrt(2) - in other words the square's lower vertex touching the arc is at a lenght of Xsqrt(2) away from the vertical radius, so by definition Cos(theta) = Xsqrt(2) I noted that Cos(theta) = 2Sin(theta) I rearranged this equation, utilizing the fact that Cos2(x) = 1 - Sin2(x), and got Sin(theta) = 1/sqrt(5) But earlier I showed Sin(theta) equals Xsqrt(2)/2. So I had the equation 1/sqrt(5) = Xsqrt(2)/2 Solving for X, the length of the square, gets you sqrt(10)/5 Square this for the area: 2/5 Complicated, but interesting and it works! :) Lots of fun

@stevenz933 2 жыл бұрын

Using a simple formula from geometry for the length of a chord = 2*(radius)*sin(θ/2), where θ is simply the angle formed from drawing two radii to the beginning and end of the chord. Using the formula, the length of the chord is Sqrt(10) / 5 and therefore the area of the square is 10/25 or 0.4.

@theredturtle4471 2 жыл бұрын

I did the method of that the squares has a point on the radii (0,a) it has a point on the circle at (2a,a). If the equation for the circle is y-sqrt(1-x^2), a=sqrt(1-4a^2), a=1/sqrt(5). The side length of the circle is sqrt2(a), so the area is 2a^2, The are is 2/5.

@ParadoxProblems Жыл бұрын

One can construct a rectangle by moving the top half of the square down to fill the gaps. The rectangle has diagonal length L=r=1 Then you take the sides of the rectangle to be x and 2x yielding Sqrt(5)*x=1 The area of the rectangle is therefore 2x*x=2/5

@Abhi-sj4wg 2 жыл бұрын

I also found smtg...take tha length of point touching radii of quater from its centre as x...now exploit symmetry of figure and mark the point of square and arc's intersection as (2x;x)...satisfy it with circles eqn...get ur x and finally the are is 0.4

@sherlock69 2 жыл бұрын

Love from India🇮🇳 ❤❤

@charlesbromberick4247 2 жыл бұрын

My survey of 1000 12 year olds indicated that exactly 0 could solve this problem (but I wouldn´t have been surprised if Richard Feynmann could have solved it at 12 years of age). I did it a slightly different way, but at the age of 12 I would have just said, "Uhhh?".

@themidnightcello8384 2 жыл бұрын

If you take the equation of the circle to be y = (1 - x^2)^0.5 and the upper diagonal side of the square to be a straight line of gradient 1 (45 degrees) y intercept at "a", you have the linear equation as y = x + a. Equating these two and substituting x = a to find the top intercept between the line and the circle leads to a simple solve-for-a which gets "a" = 1/root5. The side of the square is thus root2 times a, giving an area of 2/5

@quigonkenny 4 ай бұрын

Draw a radius to the top vertex of rhe square. Draw a diagonal down from that vertex to the vertex at the baseline. Call the diagonal length d, and the vertex on the baseline will be d/2. (d/2)² + d² = r² d²/4 + d² = r² d² + 4d² = 4r² 5d² = 4(1) d² = 4/5 Area of a square in terms of the diagonal is d²/2: d²/2 = (4/5)/2 = 2/5

@pirateskeleton7828 2 жыл бұрын

I just used t= arctan(1/2) because the vertices at which the square touches the the curve of the circle seemed to be at a 2:1 height v width (and viceversa) so I figured the area of the square is also equal to the the multiplication of sin(t) and cos(t) which yielded 0.4.

@wombat4191 2 жыл бұрын

Before watching the solution, I solved it like this. Sorry if there is some weird wording, I never did much maths in English and had to google some of the words. Name the edge of the square x. Make a right triangle so that the hypotenuse is from the tip of the quartercircle to the lower one of square's vertices at the arc of the quartercircle (so the hypotenuse has a length of 1), and the triangle's longer catethus goes along the bottom leg of the semicircle. The square can be proven to be at a 45° angle to the semicicle's legs by symmetry, so the right triangle's two catethi are sqrt(2)x and x/sqrt(2). Use the pythagoras theorem to get (sqrt(2)x)^2 + (x/sqrt(2))^2 = 1^2 2x^2 + (x^2)/2 = (5x^2)/2 = 1 x^2 = 2/5 And there is no need to go further because x^2 is the area that was asked. The problem looked a bit intimidating at first, but in the end it was just a bit of basic rules of trigonometry and the pythagoras theorem. Not sure if I could have solved this when I was 12 though.

@wombat4191 2 жыл бұрын

Oh damn the solution in the video seems more complicated. Well, I just glossed over the proof of the 45° angle and used that to make out the dimensions of the right triangle. But overall I feel like it was the simpler one, and it kinda just gave the answer accidentally in the middle of the calculations :D

@lakshyaprakash2428 2 жыл бұрын

i used coordinate consider the points on the base of quarter circle to be (a,0) and other point on the quarter circle to be (0,a).now the slope is -1. so the other oint on the quarter circle with the point on the base have a slope 1.so that other point would be (2a,a). now this point with origin as centre has length 1. so,2a^2+a^2=1. a^@=1/5, and area of the square would be (root2a)^2=2a^2=2/5

@ktomeir 2 жыл бұрын

I started by constructing the radius, and adding a line perpendicular to the left radius. this creates a right triangle that we can notice that has hypotenuse equal to one and legs equal to the diagonal of the square and half of it. By trigonometry or by Pythagoras we can calculate the diagonal and therefore the side of the square. final equation: (sin(atan(0.5))*2^(1/2))^2 = 0.4

@sage5296 2 жыл бұрын

if you look at the distance from the center of the square to the corner, you can make a right triangle with 2x and x as sides, x is root 0.2, and so the side length is root 2 times that, or root 0.4, and the area is 0.4

@PenandPaperScience 2 жыл бұрын

I love these kinds of puzzles :)

@maxair95 2 жыл бұрын

You can also divide the radius into half and calculate the hypothenus of the triangle to find the value of 1 side of the square.

@bienvenidos9360 Жыл бұрын

I solved it by drawing the radius to one of the square's corners that meets the circumference. Draw a vertical line to the bottom line of the quarter circle to make a right triangle. The radius = hypoteneuse and one of the legs = 1/2 the diagonal of the square, the other leg = the diagonal of the square. So we have a 1-2-sqrt5 triangle. Radius/diagonal= hypoteneuse/longer side. 1/x = sqrt5/2 x = 2/sqrt5 = diagonal. Side of square = diagonal/sqrt2 = 2/sqrt10 Area of square = (2/sqrt10)² = 4/10 = 0.4 units²

@klh6729 2 жыл бұрын

formula for a circle = x^2 + y^2 = r^2 in this case, x^2 + y^2 = 1 let S= side length of square for the coordinates at top of square, x= S*cos(45) or S*sqrt(2)/2 y= 2*S*sin(45) or S*sqrt(2) (S*sqrt(2)/2)^2 + (S*sqrt(2))^2 = 1 1/2*S^2 + 2*S^2 = 1 5/2S^2 = 1 S^2 = area of square = 2/5 = 0.4

@MrDcwithrow 2 жыл бұрын

Define the center of the square as point O (where the diagonals cross) having coordinates of (x,x), then the right vertex has coordinates of (2x,x), plug that into the equation of the circle, (2x)^2 + x^2 = 1^2 so 4x^2 + x^2 = 1, 5x^2 = 1, x^2 = 1/5. The side of the square squared is x^2 + x^2 (Pythagoras, with the half diagonals of the square), therefore side squared is 2/5 which is the area of the square.

@jakehobrath7721 2 жыл бұрын

So this problem made me go crazy, I attempted to solve it differently and come up with different but very close answers. I used perpendicular chord theorem, 4R^2 = x^2 + y^2. To get x and y, I determined the two chords to be the vertical diagonal thru the center of the square (times 2 of course), and a perpendicular horizontal chord from the center line to the top most point of the square both meeting at this point on the circle. Now, if we call the side length of the square a, the length from top to the highest point of square on circle b, and the diagonal of the square c, you get several equations. From Pythagorean 2b^2 = a^2 2a^2 = c^2 Perp chord theorem 4=(2c)^2 + (2b)^2 =4c^2 + 4b^2 A little substitution and you find b to be sqrt(2 - sqrt(3)). Some more substitution and you find a to be 2*(2 - sqrt(3)). To get area of the square you square it and get about .3 for the area. It’s driving me nuts what I’m doing wrong here. Please help!

@bedofsunflowers1194 Жыл бұрын

It reminded me of my secondary school math and how I have completely forgotten all of them 😂 thanks so much for your video, time to do some maths to pick them up again!

@binpepin 2 жыл бұрын

I used an equation of circle x^2+y^2=1. Then I substituted x and y with coordinates of the rightmost vertex of the square [a.sqrt(2),a.sqrt(2)/2], where a is the side of the square.

@prosperitystar 2 жыл бұрын

When i saw the video, i tried to calculate it before even watching. Here how i did: i saw the fourth part of a circle, then just drew the same square three more times and then closed them in circle. Then i drew a line from the tips to tips to make an ideal square which was equal to diameter of a square. Then i saw that the line pierces through the x line right in the middle, and also that point divides the bigger square’s sidelines to halves. I drew a hypotenuse(the legs of a triangle where 1 and 2) and the hepotenuse was equal to 2.5 of the x. Then i just divided the result by it

@mikezilberbrand1663 2 жыл бұрын

The top corner of the square has x= a/sqrt2, y=a*sqrt2. sum of the squares is 1. Solving for a^2 get 0.4. That is the area.

@wabc2336 Жыл бұрын

Since the side next to the arc is a chord, its perpendicular bisector (call it Line 1) must go to the center of the circle. Since sides of a square are parallel, Line 1 also bisects the opposite side. Therefore the triangle on the bottom left of the circle can be divided into two right triangles (T1 and T2) by Line 1. Since Line 1 also bisects the 90° angle at the bottom left of the circle, it follows that T1 and T2 each have a 90°/2 = 45° angle, making them 45-45-90° triangles. Now look at the bigger triangle T3 which consists of T1 and T2. Let its smaller side have length x so that the sides of the quarter circle are divided into x and 1-x. Let O be the center of the triangle, A be the right corner of the square, and Z be the point directly under A on the quarter circle's side. Let B be the bottom corner of the square. The angle ABZ is 45° because the square's angle is 90° and the other angle is 45°. So the right triangle ABZ is a 45-45-90° triangle. Its hypotenuse AB is equal to the hypotenuse of T3 (length: x√2) so it is congruent to T3. This makes AZ= BZ = x = OB. So OZ = 2x, AZ = x. From this we look at the triangle AOZ and find it has sides x and 2x meaning that the hypotenuse OA is x√5. And OA is the radius of the circle so x√5 = 1. So x=1/√5. The square's sides are x√2. So the area of the square is (√2/√5)² = 2/5 = 0.4. The area of the quarter circle is π/4 which is between 0.75 and 1. The square makes up 0.4π/4 = π/10 of the quarter circle, which generalizes this for any r so that the area of the square = πr²/10.

@toasteduranium Жыл бұрын

I actually saw a water fountain peeking out of a quarter circle in the corner of a hallway in a building (the University of the Incarnate Word) beneath which was a square tile that exactly modeled this situation.

@andrejsasd8904 2 жыл бұрын

I did no formulas, just approximation. It looked like the square fills about half of the area= half of the quarter of the circle, so, Pi*r^2/8. If r=1, area of square = Pi/8= 0.39. Pretty close. :D

@mathewpv681 2 жыл бұрын

If we draw the square in the adjacent quadrants of the circle. We get a right angle in the circle. Which means the hypotenuse is the diameter. One leg is x(side of square) other leg is 3x. Hence by Pythagoras we get x*x=0.4. Could do it in mind.

@smchoi9948 2 жыл бұрын

A co-geom. trial: place the sector on the Cartesian plane where the centre of the completed circle is at the origin and the straight boundaries lie on the positive sides of the axes, so x²+y²=1 is the equation of the said circle. If s is the length of the square's side, using Pythagoras' thm. twice (over right-angled isosceles △s) we see that (s/√(2),√(2) s) is where the arc meets the top tip of the square, so s²/2+2s²=1, i.e. s²=2/5 (the area required).

@sanattaori6209 2 жыл бұрын

Get approximate solution, 1/4 area of circle - three times area of triangle, 1/4 * 3.14 => 0.785−(3×(0.5×0.5×0.5)) = 0.41

@reubenbenjamin8128 2 жыл бұрын

At first I thought the solution was hard and thought of watching the video, but I quickly imagined to draw the radius to one of the square's sides which solves the problem, using simple similar triangle ratios

@jeanf6295 2 жыл бұрын

did it with trigonometry : rotate the the figure sideway 45° to the right, consider the point of the square that is one the circle and above the x-axis, that point is at an angle a, the dimensions of the square can be expressed as 2sin(a) for the height, and cos(a)-sin(a) for the length, with cos(a)-sin(a) = 2sin(a). That equation can be put under the form sqrt(10)(1/sqrt(10)cos(a)-3/sqrt(10)sin(a)) = 0, for b = Arcsin(1/sqrt(10)) = Arccos(3/sqrt(10)) this is equivalent to sin(b)cos(a)-cos(b)sin(a) = sin(b-a) = 0 Thus a = b and the square side length is equal to 2/sqrt(10), and the area to 2/5.

@derekjc777 2 жыл бұрын

I got the answer using a right angled triangle between the circle centre, the right-most square corner, and the horizontal, with side lengths of root 2 x, (root 2)x/2 and 1, where x is the length a side of the square. Pythagoras gives 1 = 2x^2 + 1/2 x^2, making x^2 = 2/5 or 0.4. A simpler solution in my opinion.

@siddharthsabharishankar7714 2 жыл бұрын

I did the same!

@turel528 Жыл бұрын

I constructed a full circle and deduced that hypotenuse of a triangle with cathetus of 1 side of square and cathetus of 3 sides of a square equal to 2. Thus, √(x^2+9x^2)=2 → 10x^2=4 → x^2 = 0.4, where x is a side of square

@NirousPlayers 2 жыл бұрын

I did another way: First i noted that the botom left triangle is an isoceles. Let's call L the side of the square and C each other side of the triange, we have using Pythagora's Theorem that L² = 2C² Now i noticed the following, We trace a line down from the right point of the square perpendicular to the horizontal line of the semi-cricle, we have then a triangle with sides L, C, H. Now we can also trace a line from the "center of what would be the circle" to the right point of the square, having another triangle with sides 1, 2C, H. Using Pythagora's Theorem we have L² = C² + H² 1² = 4C² + H² Subtract 1 from 2, we have 1 - L² = 3C² Using what we foind that L² = 2C², we have that 1 - 2C² = 3C² 1 = 5C² C² = 1/5 We now that the area of the square is L², so it is 2C². That way, the final area is A = 2/5 = 0.4

@leif1075 2 жыл бұрын

How do you bottom left is osceles..you have to prove it..did you prove the way Presh did or ankther way?

@DavidGarcia-mz5qq 2 жыл бұрын

@@leif1075 maybe using the perpendicular bisector theorem

@NirousPlayers 2 жыл бұрын

@@leif1075 Well, the square is inscribed in a quarter circle, and there is only one way to fit it so all the points touch the quarter circle. If you fold the quarter circle in half (bisecting it), the square will become one retangle, and so two sides of the bottom left triangle will join into one with equal length. Edit: Correction, there are actually two ways the square can have all 4 points touching the quarter circle. The other one is if it has two of it sides parallel to two of the straight lines of the quarter circle.

@spafon7799 2 жыл бұрын

Can also use intersecting chords which gives (1+3x)*(1-3x)=x^2, thus 1-9x^2=x^2 or 1=10x^2 thus x^2=1/10 and (2x)^=4/10.

@timeonly1401 Жыл бұрын

I applied the law of cosines to the triangle of sides s, s/sqrt(2) and 1, with the angle opposite side of length 1 being 135 degrees. Then solved for s^2, which is the area of the square: 1^2 = s^2 + s^2/2 -2[s^2/sqrt(2)]cos(135) . Since cos(135) = -1/sqrt(2), we have: 1 = s^2 ( 1 + 1/2 + 1) = (5/2) s^2. So that: s^2 = 2/5 = Area of square. Done.

@surplus3034 2 жыл бұрын

as mentioned by "SenselessUsername" yesterday,assuming each side of square = a . place the quarter circle on a coordinate system with center at 0,0 results in 2 corner of the square on the circle to be (√2.a, a/√2) and (a/√2, √2.a) .plug either in the unit circle formula x^2+y^2=1 and you'll have the result.

@panlomito 2 жыл бұрын

My way to the answer: side of the square = x, left triangle a-a-x mirrored over the right point. Then (2a)² + a² = 1² or a² = 1/5 and Area square = x² = a² + a² = 2/5 = 0,40... easy.

@richardrhodes5644 2 жыл бұрын

Mine too. Seemed simpler and quicker than solution presented by Presh. 5 lines of algebra using his "favorite" formula!

@jaypokale35 Жыл бұрын

Draw complete circle and mirror the square along any axis then use chord intersection theorem: --> (1 - x√2)(1 + x√2) = (x/√2)² --> 1 - 2x² = x²/2 --> x² = 2/5 = 0.4

@user-xq9dn9ri6i 2 жыл бұрын

what a beautiful solution it is

@Ozymandi_as Жыл бұрын

Just consider horizontal diagonal of the square, which must hypotenuse of right triangle with sides in ratio 2s:1s, length 1. Therefore s = √(⅕), therefore S² = 2s² = ⅖.

@0over0 2 жыл бұрын

Consider R Tri made of the 2 horizontal corners of box, and center of circle. Solve for D^2, square's diagonal squared: D^2 + (1/2 D)^2 = 1^2. D^2 = 4/5. Let S= square's side len. Since S = D/sqrt(2), S^2 = D^2/2 = 2/5.

@hameedamathtuber 2 жыл бұрын

That was really super cool

@jayantaboral6446 2 жыл бұрын

One can solve this with the help of coordinate geometry also The vertices are (a,0), (2a,a), (a,2a) and (0,a). Now any one of the points, (a,2a) and (2a,a) will lie on the circle x^2+y^2=1, that gives us (2a)^2+a^2=1 or 5a^2=1. That gives a=1/√5. Each of the sides of the square is a√2=√2/√5. So area of the square = (√2/√5)^2 = 2/5 = 0.4

@francesco8422 2 жыл бұрын

The eastern way to solve is to consider that the square extremities cuts the radius of the 1/4 circle in the middle so you apply the Pitagora theorem to find a line and then you multiply it by itself to find the solution

@altheengineer9632 2 жыл бұрын

I think I found a different way to solve it, by I am not sure, I got a same answer though. I extended the circle so it is 4/4 of circle instead of 1/4. The made congruent square on the other 3/4 of the original drawing. Then, the diamater is 2, so I extend the line to connect one vertices of the newly formed rectangle. So 2 is now a diagonal. Now, let’s name x one side of a square. Because there are 3 total congruent squares to make a rectangle, its length and width is 3x and x. Using pythag, we find that 4 = 10x^2, simplify and get x = 2/rad 10. Now, to get one area of square, square X. Now, it is 4/10 or .4 units. I am also in the age group, and this problem wasn’t that challenging, but then again, I skipped a few grades in math.

@youknowme3346 2 жыл бұрын

Amazing solution

@Sahil-my9sv 2 жыл бұрын

I solve by another way, join center of circle to one of squares corner, and it's length is 1, now apply cosine rule, and get side of square as √(2/5)

@kingcrimson1631 2 жыл бұрын

if the side of the square is a, then the side of the smallest triangle is a/sqrt(2). Then the rectangle that covers the bottom half of the square (diamond) has sides of length sqrt(2)a and a/sqrt(2) and diagonal of 1. Pythagoras gives 2a^2 + a^2/2 = 1^2, then a = sqrt(1/2.5) so the area is a^2 = 4/10